Problem: $\dfrac{dy}{dt}=3t^2+1$ and $y(1)=5$. What is $t$ when $y=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $t=0$ (Choice B) B $t=\dfrac32$ (Choice C) C $t=1$ (Choice D) D $t=-\dfrac32$ (Choice E) E $t=-1$
Explanation: The differential equation is separable. What does it look like after we separate the variables? $dy=\left(3t^2+1\right) dt$ Let's integrate both sides of the equation. $\int dy=\int\left(3t^2+1\right) dt$ What do we get? $y=t^3+t+C$ What value of $C$ satisfies the initial condition $y(1)=5$ ? Let's substitute $t=1$ and $y=5$ into the equation and solve for $C$. $\begin{aligned} 5&=1^3+1+C\\ \\ C&=3 \end{aligned}$ Now use this value of $C$ to find $t$ when $y=3$. $\begin{aligned} 3&=t^3+t+3\\ \\ 0&=t^3+t\\ \\ 0&={\cancel{(t^2+1)}}\cdot t\\ \\ t&=0 \end{aligned}$